HDU 3018 Ant Trip(欧拉回路一笔画问题)

题意：给一个无向图，N个顶点和M条边，问至少需要几笔才能把所有边画一遍

思路：只要知道一个结论就随便做了。

如果该连通分量是一个孤立的点,显然只需要0笔.

如果该连通分量是一个欧拉图或半欧拉图,只需要1笔.

非(半)欧拉图需要的笔数==该图中奇数度的点数目/2

对于每个以i为根的连通分量我们记录属于该连通分量的点数目num[i]和该连通分量中奇度点的个数odd[i]，然后瞎搞就好了

#include <cstdio>
#include <queue>
#include <cstring>
#include <iostream>
#include <cstdlib>
#include <algorithm>
#include <vector>
#include <map>
#include <string>
#include <set>
#include <ctime>
#include <cmath>
#include <cctype>
using namespace std;
#define maxn 100000+100
#define LL long long
int cas=1,T;
int pre[maxn];
int Find(int x)
{
return pre[x]==-1?x:pre[x]=Find(pre[x]);
}
int n,m;
int num[maxn];
int odd[maxn];
int degree[maxn];
int main()
{
while (scanf("%d%d",&n,&m)!=EOF)
{
memset(pre,-1,sizeof(pre));
memset(num,0,sizeof(num));
memset(odd,0,sizeof(odd));
memset(degree,0,sizeof(degree));
for (int i = 0;i<m;i++)
{
int u,v;
scanf("%d%d",&u,&v);
degree[u]++;
degree[v]++;
u = Find(u);
v = Find(v);
if (u!=v)
pre[u]=v;
}
for (int i = 1;i<=n;i++)
{
num[Find(i)]++;
if (degree[i]%2)
odd[Find(i)]++;
}
int ans = 0;
for (int i = 1;i<=n;i++)
{
if (num[i]<=1)
continue;
else if (odd[i]==0)
ans++;
else if (odd[i]>0)
ans+=odd[i]/2;
}
printf("%d\n",ans);
}
//freopen("in","r",stdin);
//scanf("%d",&T);
//printf("time=%.3lf",(double)clock()/CLOCKS_PER_SEC);
return 0;
}

题目

Description

Ant Country consist of N towns.There are M roads connecting the towns.

Ant Tony,together with his friends,wants to go through every part of the country.

They intend to visit every road , and every road must be visited for exact one time.However,it may be a mission impossible for only one group of people.So they are trying to divide all the people into several groups,and each may start at different town.Now
tony wants to know what is the least groups of ants that needs to form to achieve their goal.



Input

Input contains multiple cases.Test cases are separated by several blank lines. Each test case starts with two integer N(1<=N<=100000),M(0<=M<=200000),indicating that there are N towns and M roads in Ant Country.Followed by M lines,each
line contains two integers a,b,(1<=a,b<=N) indicating that there is a road connecting town a and town b.No two roads will be the same,and there is no road connecting the same town.

Output

For each test case ,output the least groups that needs to form to achieve their goal.

Sample Input

3 3
1 2
2 3
1 3

4 2
1 2
3 4

Sample Output

1
2

Hint

New ~~~ Notice: if there are no road connecting one town ,tony may forget about the town. In sample 1,tony and his friends just form one group,they can start at either town 1,2,or 3. In sample 2,tony and his friends must form two group.